2022-03-15 19:59:12 +01:00
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# Impedance/Reactance of capacitors
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## Capacitive Reactance
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Is a measure of a capacitors opposition to alternating current.
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$Xc$ in $\ohm$
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$X_{c} = \frac{1}{2 \pi fC}$
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$Xc = \textit{Capacity in } \ohm$
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f = Frequency in Hertz
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C = Capacitance in Farads
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2022-04-15 14:51:51 +02:00
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![](graphXc.gif)
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2022-03-15 19:59:12 +01:00
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Higher Frequence $\Rightarrow$ Lower Current Flow
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Higher Capacitance $\Rightarrow$ Lower Current Flow
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When the Frequency is 0, the capacitor acts as an open circuit
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When the Frequency is really high, the capacitor is equal to a simple wire
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**Example:**
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Calculate the capacitive reactance of a 220nF capacitor at a frequency of 1kHz and 20kHz
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$$
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\begin{flalign}
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&X_{c} = \frac{1}{2 \pi * 1000 * 220 * 10^{-9} } \\
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&X_{x} \approx \textbf{723.43} \ohm\\
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\\
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&X_{c} = \frac{1}{2 \pi * 20000 * 220 * 10^{-9} } \\
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&X_{x} \approx \textbf{36.17} \ohm\\
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\end{flalign}
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$$
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Here we can see when the frequency increases the reactive capacitance decreases
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**Example 2:**
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```circuitjs
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$ 1 0.000005 10.20027730826997 50 5 43 5e-11
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v 208 256 208 144 0 1 80 5 0 0 0.5
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r 208 144 336 144 0 100
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c 336 144 336 256 0 0.000029999999999999997 -2.4446139526159825 0.001
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w 336 256 208 256 0
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```
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How would we calculate the $I_{rms}$ of this circuit, we'll basically using Ohms Formular
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$$
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I_{rms} = \frac{V_{rms}}{R1+X_{c}}
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$$
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The Problem is, we can't just simply add up R1 and Xc, because Xc is shifted by 90°. We need to add them up as Vectors:
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$$
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2022-03-20 23:15:38 +01:00
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R_{e} = \sqrt{R1^2+X_{c}^2}
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2022-03-15 19:59:12 +01:00
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$$
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Lets fill in the numbers from the circuit above and test it out:
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$$
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\begin{flalign}
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&X_{c} = \frac{1}{2 \pi * 80 * 30 * 10^{-6}} &&\\\
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&X_{c} \approx 66.3 \ohm \\
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&V_{rms} = 3.5v \\
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\\
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&I_{rms} = \frac{3.5}{\sqrt{100^2+66.3^2}} \\
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&I_{rms} = \frac{3.5}{119.98} \\
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&I_{rms} = 0.029171033 A \\
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&I_{rms} \approx 29.17mA
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\end{flalign}
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$$
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## Reality
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In reality capacitors are not perfect, they are more like:
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2022-04-15 14:51:51 +02:00
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![](rlc-capacitor.svg)
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2022-03-15 19:59:12 +01:00
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So the have a $ESR$ and $X_{C}$ and $X_{L} / ESL$
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$$
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C_{IMP} = ESR + X_{C} + X_{L}
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$$
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Due to this the frequency to impedance curve of real capacitors look something like this.
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2022-04-15 14:51:51 +02:00
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![](EMC-9_graf_01.gif)
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2022-03-15 19:59:12 +01:00
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When we add multiple capacitors we can get a curve looking like this
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2022-04-15 14:51:51 +02:00
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![](rlc-capacitor-multiple.png)
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